Simplify; express your answer in exponential form. Assume $r\neq 0, n\neq 0$. $\dfrac{{(r^{-4})^{-5}}}{{(r^{-2}n^{-2})^{-2}}}$
Answer: To start, try working on the numerator and the denominator independently. In the numerator, we have ${r^{-4}}$ to the exponent ${-5}$ . Now ${-4 \times -5 = 20}$ , so ${(r^{-4})^{-5} = r^{20}}$ In the denominator, we can use the distributive property of exponents. ${(r^{-2}n^{-2})^{-2} = (r^{-2})^{-2}(n^{-2})^{-2}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(r^{-4})^{-5}}}{{(r^{-2}n^{-2})^{-2}}} = \dfrac{{r^{20}}}{{r^{4}n^{4}}}$ Break up the equation by variable and simplify. $\dfrac{{r^{20}}}{{r^{4}n^{4}}} = \dfrac{{r^{20}}}{{r^{4}}} \cdot \dfrac{{1}}{{n^{4}}} = r^{{20} - {4}} \cdot n^{- {4}} = r^{16}n^{-4}$.